Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{-r - 7}{-3r^2 - 6r + 105} \times \dfrac{r^2 - 1}{r + 1} $
First factor out any common factors. $q = \dfrac{-(r + 7)}{-3(r^2 + 2r - 35)} \times \dfrac{r^2 - 1}{r + 1} $ Then factor the quadratic expressions. $q = \dfrac {-(r + 7)} {-3(r + 7)(r - 5)} \times \dfrac {(r + 1)(r - 1)} {r + 1} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {-(r + 7) \times (r + 1)(r - 1) } { -3(r + 7)(r - 5) \times (r + 1)} $ $q = \dfrac {-(r + 1)(r - 1)(r + 7)} {-3(r + 7)(r - 5)(r + 1)} $ Notice that $(r + 7)$ and $(r + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-(r + 1)(r - 1)\cancel{(r + 7)}} {-3\cancel{(r + 7)}(r - 5)(r + 1)} $ We are dividing by $r + 7$ , so $r + 7 \neq 0$ Therefore, $r \neq -7$ $q = \dfrac {-\cancel{(r + 1)}(r - 1)\cancel{(r + 7)}} {-3\cancel{(r + 7)}(r - 5)\cancel{(r + 1)}} $ We are dividing by $r + 1$ , so $r + 1 \neq 0$ Therefore, $r \neq -1$ $q = \dfrac {-(r - 1)} {-3(r - 5)} $ $ q = \dfrac{r - 1}{3(r - 5)}; r \neq -7; r \neq -1 $